Peter Williams

GCE Chemistry 6CH05/01 “General Principles of Chemistry II

Transition Metals and Organic Chemistry” Electrophilic substitution (aromatic)

i)

Attacking species formation; (if applicable) Electron pair (curly arrow) from ring to positively charged; second carbon of carbocation; Structure of intermediate must include positive sign; Electron pair from C­H bond reforms delocalized ring;

ii) Naming reaction: Nucleophilic substitution

Nucleophilic Sn1 and Sn2 + prediction of stereochemistry of molecule (from mechanism) Using an SN1 mechanism:

selection of a chiral starting material; curly arrow from C­X bond to X; and intermediate carbocation; curly arrow from nucleophile (can come from negative charge) (1) planar intermediate attacked from either side to give a racemic mixture

OR intermediate equally attacked from either side to give a racemic mixture;

Using an SN2 mechanism:

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Peter Williams

selection of a chiral starting material; curly arrow from nucleophile (can come from negative charge);

curly arrow from C­X bond to X; to give correct transition state; attack from opposite side to C­X bond gives inverted product can be shown on a diagram

Using nucleophilic addition to C= O: Selection of any aldehyde (other than methanal) or any asymmetric ketone;

Curly arrow from nucleophile (can come from negative charge) to C of C=O and curly arrow from = to O (1) Intermediate;

Arrow from O– of intermediate to H+; planar molecule attacked from either side to give a racemic mixture;

OR molecule equally attacked from either side to give a racemic mixture;

*note: Sn2 uses a transition state NOT an intermediate carbocation. Advantage of using a solid (heterogeneous) catalyst

Advantage: (Graphite/sold) catalyst easier to remove / separate / can be filtered off (from reaction mixture) / graphite can be reused;

Justification: AlCl3 is soluble or graphite is insoluble /different state / different phase; Why methyl benzenes are more reactive than ordinary benzenes

Greater electron density in ring / ring is activated / more susceptible to electrophilic attack;

Due to electron releasing / donating methyl groups; Why methyl benzenes are more reactive than ordinary benzenes 2

(Methylbenzene) is more susceptible to electrophilic attack/attack by positive species/makes it a stronger nucleophile

How methyl groups ‘activate’ the benzene ring (Methyl group) donates/increases electron density to the ring/feeds electrons into ring Allow the methyl group is electron releasing

REJECT: Donates lone pair of electrons

Ring becomes more electronegative Just ‘inductive effect’

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Peter Williams

Half equations of Methanedioic acid + manganese ions Oxidation: (COOH)2 → 2CO2 + 2H+ + 2e­; Reduction: MnO4− + 8H++ 5e− → Mn2+ + 4H2O;

Full equation from half­equation + showing ratio 5(COOH)2 + 2MnO4−+ 6H+ → 10CO2 + 2Mn2+ + 8H2O;

5:2 ratio; Writing half equations at anode + cathode (H2 and O2)

Anode: H2 − 2e(­) → 2H+; Cathode: O2 + 4H+ + 4e(­) → 2H2O;

Advantage of fuel cell with hydrogen in comparison to combustion of hydrogen One advantage e.g. quieter, more efficient (energy transfer), no NOx formed

Ignore references to carbon dioxide and / or water as only product Biomass fuels and hydrogen fuel over fossil fuels (ethanol etc.)

Ethanol can be obtained from biomass / plants / fermentation / ethanol is a bio fuel; hydrogen from (electrolysis of) water using a non­fossil source of energy; these are renewable / fossil fuels are a finite resource;

Calculation of % mass present (further titration calculations) Scenario:

i)

Moles of MnO4− = 11.30/1000 x 0.010 = 1.13 x 10−4 (mol); Moles of (COOH)2 in 10 cm3 = 1.13 x 10−4 x 5/2 = 2.825 x 10−4 (mol); Moles of (COOH)2 in whole sample = 2.825 x 10−4 x 50 = 0.01412(5) (mol); Mass of acid = 0.01412(5)x 90 = 1.27 g; % in leaves = 1.27/250 x 100 = 0.51 (%);

ii) (Percentage error in 11.3cm titre) ± 0.05 cm3; [(0.05 x 2) / 11.3] x 100 = 0.88%;

Calculation of atoms present in compound from % mass C 60(%)/12 = 5; H 8(%)/1 = 8; O 32(%)/16 = 2;

Using calculations to determine the formula of product

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Peter Williams

amount of carvone = (4.5 ÷ 150) mol = 0.03 mol (1) amount of hydrogen = (1.44 ÷ 24) mol = 0.06 mol (allow 1st mark for either of the mole calculations) so two double bonds are reduced

OR 2 moles H2 : (1 mol carvone)

OR 4 mole H : (1 mol carvone)

Carvone molecule (text) reacted with bromine for both double bonds having HBr added

ignore added hydrogens for correct orientation in exocyclic double bond

Distinguishing between compounds using bromine reaction

HBr can be eliminated using a hydrogen from the carbon on either side of the bromine;

which would then give a double bond in a different position from that in carvone;

Why results may be deemed unreliable (other than accuracy) (regarding titration) Only one titration carried out; Leaves may contain other substances that MnO4−could oxidize/ react with; Not all ethanedioic acid extracted from leaves; Different amounts of acid from different leaves;

ALLOW temperature too low / below 60˚C; Shape of complex ion

Mn2+ = octahedral; Explanation as to why transition metals are coloured in solution

(Ligands cause) d orbitals / sub­shell / sub level to split; Some frequencies of light (energy) are absorbed; To promote electrons (within d level / d → d transitions);

*note: for (1 Mark question) can just say it has partially filled d­orbitals

ALLOW as alternative for second mark: Remaining light is transmitted / reflected (resulting in the colour seen)

Characteristics of a transition metal

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Peter Williams

Variable/varying/different/several/ more than one oxidation state /number; Complex (ion formation);

Characteristics of a transition metal (another mark scheme) complex ions / complexes; coloured ions / compounds / solutions; catalytic properties; paramagnetic;

Allow: coloured complexes Ignore ‘partially filled d­orbitals’

*note: most questions regarding this will say ‘other than...’; in this case ‘other than variable oxidation states’. Best not to say ‘partially complete d­orbitals’ as this is rejected most of the time. Electron configuration of Fe3+

1s22s22p63s23p63d5(4s0); Accept [Ar]3d5(4s0); (Ion) has an incompletely filled (3d­orbital )/ sub­shell / unpaired d electron;

Explanation as to how ligands join to metal ion (Fe and Nitrogen from text) (haemoglobin example)

Lone pair(s) (from nitrogen(s); Forms dative / dative covalent / coordinate bond (with Fe2+);

Bidentate explanation Forms two (dative/covalent) bonds/has two lone pairs (to the Transition Metal/ion);

H2NCH2CH2NH2 can act as a bidentate ligand whereas H2NNH2 cannot Both have two nitrogen atoms with lone pairs or implied;

OR Far enough apart/longer chain in between in en (but not in hydrazine)/too close in

hydrazine/hydrazine is too short/not as long; Complex ions reacting to form acidic solution (with water) [Cr(H2O)6]3+ + H2O [Cr(H2O)5(OH)]2+ + H3O+

ALLOW [Cr(H2O)6]3+ + H2O [Cr(H2O)5(OH)]2+ + H2O + H+ Why 3+ complex ions have a lower pH than 2+ complex ions

The concentration of oxonium / hydrogen ions is less in the [Cu(H2O)6]2+ / fewer hydrogen ions produced or reverse argument based on Cr ion;

Because copper ion is 2+ whilst the chromium ion is 3+ / charge on copper ion is less than charge on Cr ion / less charge density on 2+ ions / Cr (3+) draws more electron density from the O­H bond;

NaOH and complex ions NaOH is a (strong) base / alkali; Cr(H2O)3(OH)3 loses (three) protons / undergoes further deprotonation;

OR Cr(OH)3 is amphoteric (so reacts with strong bases); To reverse reaction 4 add (sulfuric) acid / H+ / HCl;

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Peter Williams

Why edta ions is more stable than other complexes (including entropy) [Cr(NH3)6]3+ + (edta)4­ → [Cr(edta)]− + 6NH3; During the reaction number of particles increases (2 to 7) / more moles of product than

reactants AND entropy (of system) increases; Drawing complex ions (octahedral example)

ignore absence of charge clearly octahedral (ignore bonds to the H in H2O), but allow some

latitude in the symbols used to show the 3D structure; Wedges do not have to be exact – if used they are enough to show

3D if the axial bonds are lines; The word ‘octahedral’ does not salvage a poor drawing; dative (covalent) / coordinate (bond) not just shown by an arrow; lone pair (of electrons on the oxygen) can be shown on the

diagram; Using data to provide molecular formula

Mass of C in CO2 = 12/44 x 0.88 = 0.24 g; Mass of H in H2O = 2/18 x 0.216 = 0.024g; So mass of oxygen = 0.328 – ( 0.24 + 0.024)

= 0.064 g; Moles of C = 0.24/12 = 0.02; Moles of H = 0.024/1 = 0.024; Moles of O = 0.064/16 = 0.004; Ratio = simplest ratio = 5:6:1 so C10H12O2;

Presence of a C=C bond

Add (small amount of) Br2 / bromine; (Br2 turns from orange / yellow / red­brown to) colourless / decolourised;

OR Add (small amount of) acidified KMnO4(aq); KMnO4(aq) turns from purple/pink to colourless / brown;

Test for ­COOH Test : add NaHCO3 /Na2CO3/sodium carbonate (solution); Result: Fizzies/bubbles/large volume neutralized;

Avoiding loss of volatile material (reaction mixture) (Heat under) reflux;

Steam distillation apparatus steam source and r.b /pear­shaped flask (and clove buds);

OR r.b /pear­shaped flask being heated and containing water (and clove buds);

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Peter Williams

Condenser with water jacket, in correct position and direction of water flow; Collection vessel;

­1 if apparatus does not work e.g. sealed; ­1 for no joints or leaky joint;

Drying procedure Mix organic solvent and oil­water mixture in a separating funnel then separate; Distil / rotary evaporate (to separate clove oil from organic solvent); Add (anhydrous)CaCl2 /(anhydrous) MgSO4 /(anhydrous) Na2SO4 /silica gel / calcium

oxide to clove oil, (then filter / decant); Calculating E cell values (from the data book the equations are)

1. Cu+(aq) + e− → Cu(s) E = +0.52 (V); 2. Cu2+(aq) + e− → Cu+(aq) E = +0.15 (V); So E cell = 0.52 − 0.15 = +0.37 (V);

Units (1) value (1) *note: rules to calculate E cell (not a mark scheme):

E cell = E cathode ­ E anode Cathode= reduction Anode = oxidation Cathode= ½ equation 1 Anode= ½ equation 2

*must be positive to be spontaneous! E cell calculation + determining if a reaction is feasible using E cell values

Half­equation I: Cr3+(aq) + e–o Cr2+(aq); Half­equation II: Cr2O72–(aq) + 6e– + 14H+(aq) o 2Cr3+(aq) + 7H2O(l); Half­equation I: ­ 0.41 (V); Half­equation II: +1.33 (V); Calculate E cell= ­0.41 ­ (+1.33) = ­1.74;

Combined half equations: 8Cr3+ + 7H2O 6Cr2+ + Cr2O72­ +14H+; E cell reaction= — 1.74V; So not feasible on condition of negative value;

Determining if a reaction is feasible using E cell values 2 − 0.66(V); reaction not feasible since the potential is negative;

Setting up an E cell equation (Iron III oxide from text) (using databook) Ecell for reaction is (+) 0.84 (V) (so will work) / Ecell for item 44 is more positive than for

item 19 / illustrate using anti­clockwise rule; 2Fe + O2 + 2H2O → 2Fe2+ + 4OH−; Ecell for reaction is (+)0.96 (V) (so will work) / Ecell for item 44 is more positive than for

item 17 / illustrate using anti­clockwise rule; 4Fe(OH)2 + O2 + 2H2O → 4Fe(OH)3l;

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Peter Williams

Using equilibrium knowledge in E cell equation (example from text)

(When the hydroxide ion concentration is increased) the equilibrium (of the second half

equation) moves to the left/back; E becomes less positive/more negative/decreases/reduces; Therefore E cell becomes positive (so reaction feasible);

Why cell hydrogen­oxygen E­cell values are same in acidic/alkaline conditions Reaction / equation for both is 2H2(g) + O2(g) → 2H2O(l);

ALLOW statements such as ‘they both produce water from hydrogen and oxygen’ / ‘reactants and products are the same’;

Explanation of different colours seen when different compounds reacted (E cell values) Scenario:

Copper (formed (by disproportionation)) is oxidized (by nitric acid) [must be stated in words];

Relevant half equations are: 1. Cu2+(aq) + 2e− → Cu(s) Eo = +0.34 (V); 2. 2NO3−(aq) + 4H+(aq) + 2e− → N2O4(g) + 2H2O(l) E Cell = +0.80 (V);

OR NO3−(aq) + 3H+(aq) + 2e− → HNO2(aq) + H2O(l) E Cell = +0.94 (V) (1)

Correct overall equation scores both marks: Cu + 2 NO3− + 4H+ → Cu2+ + N2O4 + 2H2O

OR Cu + NO3− + 3H+ → Cu2+ + HNO2 + H2O (So) E Cell is +0.46 (V) (or +0.60 (V) or just ‘positive’)

Why E cell values may be different / incorrect (electrode potential) values are for standard conditions; (nitric) acid is concentrated / not 1 mol dm­3 / not 1 M;

Amphoteric explanation + example (Zinc)

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Peter Williams

zinc hydroxide/oxide amphoteric because it reacts with alkali (to give a solution of a zincate);

and reacts with acid (to give a salt); zinc hydroxide is / acts as both an acid and an alkali; hexaquazinc or hydrated zinc ions exchanged water for ammonia or other named ligand;

OR Zn(H2O)6 2+ + 4NH3 → etc; Allow any number of ammonias from 1 to 6;

Allow balanced equations, ionic or full. Ligand exchange reaction must start with a complex ion;

Explanation of disproportionation (simultaneous) oxidation and reduction;

OR Simultaneous increase or decrease in oxidation number of an element;

ALLOW ‘Species’ ‘atoms of the same type’ for ‘element’;

Explanation in terms of copper; IGNORE Atom / ion / compound / substance / reactant

Reverse disproportionation reaction (example) Two (less stable) oxidation states/one higher and one lower oxidation state of the

same/an element react to form one(more stable) oxidation state;

ALLOW The reverse reaction is a disproportionation in which (one oxidation state of) the same/an

element and it EITHER: reacts to give one higher and one lower oxidation state/two oxidation states;

OR is both oxidized and reduced;

Correct oxidation states +7 and +4 to +6; Mn(VII) and Mn(IV) to Mn(VI);

Why isomers with C=C show geometrical isomerism There is a barrier to rotation about a (C=C) bond;

ALLOW restricted / limited / no rotation ; Each carbon atom (in the C=C double bond) has (two) different atoms / groups attached;

Why some aromatic compounds can show geometrical isomerism without C=C There is a barrier to / restricted rotation about the ring;

OR The ring behaves like a double bond;

E­Z isomers Explanation of precedence/priority in terms of atomic numbers/masses of the attached

groups; OR

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Peter Williams

Highest­precedent/priority groups on each carbon are on opposite sides of the molecule (so E isomer);

Explanation of racemic mixture Ethanal is a planar molecule; Attack (from CN­ to give the cyanohydrin) is (equally likely) from either side/above or

below/from both sides (of the molecule) (so a racemic mixture is formed); Disadvantage of racemic mixture in pharmaceuticals

Receptors for the compound in the body are often stereospecific so only one stereoisomer is pharmacologically active;

OR Body recognises one (stereo)isomer;

ALLOW: Only one (stereo)isomer is active OR

One/the other isomer may be toxic/dangerous/harmful; OR

One isomer destroys body cells; OR

(Different) isomers have different biological/pharmacological/biochemical properties; Disadvantage of racemic mixture in pharmaceuticals 2

there is no racemisation so the product will not be a mixture; no need to separate (the enantiomers); do not have to discard an unwanted enantiomer / atom economy is higher;

Compound to zwitterion (isoelectric point) i)

Formula showing –NH3+ and —COO—/—CO2—; Charges can be anywhere on functional group

ii) (polymer of zwitterion molecule) Correct peptide link; Minimum two residues and extension to the rest of the

molecule; Melting point of zwitterion

High energy needed (to overcome)strong ionic/electrostatic forces;

OR strong forces between oppositely charged ions/between positive and negative; between different (zwitter)ions;

OR between –NH3+ and —COO—;

Identifying unknown amino acids KP1 Spot (of hydrolysate) on paper/tlc/thin layer chromatogram; Mark spots of (known) amino­acids/measure Rf; Run in (suitable) solvent/discussion of comparative solubilities; (Spray with) ninhydrin (and heat);

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Peter Williams

Marker spots and the unknown spots correspond; Benzene: proving there are no double bonds

Double bonds expected to react with bromine water turning it colourless; OR

Bromine water remained yellow/orange/red/brown; So benzene does not contain double bonds;

Why enthalpy change of hydrogenisation for benzene is less exoythermic than expected The p/pi­/∏/6 electrons (of carbon);

OR ∏ system; Electrons are delocalised around the ring; Which gives the molecule greater stability/need more energy to break the bonds in

benzene (and hence a less exothermic hydrogenation enthalpy); Equation for formation of electrophile (AlCl3 catalyst)

CH3Cl + AlCl3 → CH3+ + AlCl4− Allow CH3δ+ AlCl4 δ­ and other halogens

Ignore state symbols and curly arrows Intermediate (benzene ring intermediate)

Ignore curly arrows and use of wedges/dashes; Ignore attempts to complete mechanism if intermediate is correct;

Must show reasonable delocalisation over at least 3 carbon atoms Allow positive charge anywhere inside benzene ring;

Reject: upside­down delocalization Why benzene undergoes substitution rather than addition reactions

The delocalization of the (π) electrons of the ring make benzene more stable (than 1,3,5­ cyclohexatriene);

Substitution retains this (stable) arrangement; OR

Addition removes this (stable) arrangement; Benzene to benzaldehyde (substitution­electrophilic)

Formation of electrophile (curly arrow, structural formulae not required). [Positive charge may be anywhere on the electrophile];

ALLOW HCl + CO for HCOCl; ALLOW Non­displayed electrophile;

Curly arrow from benzene ring to electrophile; Wheland structure with gap opposite tetrahedral carbon; Curly arrow from C―H bond into ring and formation of correct

organic product; OR

Kekulé structures;

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Peter Williams

First curly arrow may come from any part of the delocalisation circle Second curly arrow may come from any part of the C–H bond Positive charge on the Wheland structure may be in any part of the horseshoe

Why phenol may react faster than phenyl methanol(regarding electrophilic substitution) Electron density of the ring increased Due to donation of oxygen / OH group lone pair to the ring (in phenol) oxygen / OH group attached directly to ring Oxygen / OH group in phenylmethanol too far away / not attached directly to ring (In phenol) lone pair overlaps with the π electrons / delocalised electrons (of the ring);

Why phenol can be nitrated under milder conditions than benzene (lone pair of) electrons on the oxygen atom or on the OH group is delocalised /

incorporated into the ring OR

the OH group is electron donating; so the ring in phenol is more negative / has increased electron density / ring is more

nucleophilic / hence more susceptible to electrophilic attack; OR

the OH group activates the ring; Electrophilic substitution of benzene (via friedel­crafts ACYLATION) Formation of electrophile: C6H5COCl + AlCl3 → C6H5CO+ + AlCl4­

First curly arrow, as shown, to start from inside the hexagon to the correct C+ carbon (i.e. not to the benzene ring) Note the + must be on the C of the C=O/CO;

Intermediate correctly drawn; Second curly arrow as shown from C—H bond to reform the ring, not from the H atom in

this bond; Nitration of benzene (substitution­electrophilic)

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Peter Williams

i) Formation of electrophile: H2SO4 + HNO3 → NO2+ + H2O + HSO4­;

OR H2SO4 + HNO3 → H2NO3+ + HSO4;

ii) Electrophilic substitution reaction: arrow showing attack on the nitronium ion

with arrow going to N atom, or into the C – N gap;

Arrow must start at or inside ring; Ignore position of + charge

structure of the intermediate showing reasonable delocalisation (over at least 3 carbon atoms);

arrow from the bond showing the loss of H+from the intermediate;

Removal by hydrogen sulphate ion preferable but not essential; Kekulé structures score full marks.

Benzene to azo dye (substitution­electrophilic)

i) Formation of electrophile:

2H2SO4 + HNO3 → NO2+ + H3O+ + 2HSO4—;

ii) Mechanism:

Attack on nitronium ion arrow must start on or in the benzene; Wheland intermediate: Can be a part, but not complete circle, in correct place inside

ring BUT part circle must cover minimum of 3 carbon atoms AND must not include where nitro group is attached and must positive charge somewhere;

Either but only one of first two marks can be lost if bond is clearly to oxygen;

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Peter Williams

Arrow from H bond into the ring to produce either H+or H2SO4 and return to aromaticity; Benzenediazonium (cat)ion (salt included)

Positive charge can be on either N; Cl­ may be given as well;

ALLOW circle in benzene ring and hydrogens/carbons displayed; *note: including Cl­ not always given mark when asked to draw diazonium ion but never penalised Benzenediazonium cation to azo dye

Add phenol in sodium hydroxide/OH−/alkali; Correct structure for the ­N=N­ bond between 2 benzene rings; Remainder of molecule;

Bromobenzene via halogen carrier Attacking species formation:

Br2 + FeBr3 → FeBr4­ + Br+

Arrow from benzene ring electrons (from inside the hexagon) to Br+/Brδ+(…..Brδ­FeBr3);

Correctly drawn intermediate with delocalization; covering at least three carbon atoms, but not the carbon atom bonded to the bromine

with the positive charge shown inside the hexagon; The bonds to H and Br may be dotted; Arrow from or close to bond to H to centre of ring and H+/HBr as a product;

Ideal temperature of a given reaction (temps from text) < 0o C/temperature too low: reaction too slow/insufficient energy to overcome activation

energy; > 10o C/temperature too high: diazonium ion decomposes/produces phenol;

Suggesting solubility of a compounds (azo dye from text)

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Peter Williams

—SO3— are solvated / hydrated; Can be drawn with polar H of water;

OR Negative ion bonds with/attracted to water; Nitrogen/oxygen atoms hydrogen­bonded (to water); Can be drawn;

Why aromatic compounds are only slightly soluble in water (even though some may form hydrogen bonds)

Net forces between paracetamol and water are less than the forces between water and water and / or paracetamol and paracetamol

Allow benzene / ring doesn’t interact with water; Allow benzene ring is hydrophobic / non polar / only forms London forces / can’t form

hydrogen bonds; Percentage mass calculation

Moles S2O32− = 27.85 x 10−3 x 0.0631 (1) (= 1.757335 x 10−3); Moles of I2 remaining = Moles S2O32− ÷ 2 = 27.85 x 10−3 x 0.0631 ÷ 2 = 8.786675 x

10−4 = 8.79 x 10−4; Moles ascorbic acid = moles I2 at start − Moles I2 remaining = 2.00 x 10−3 − 8.786675

x 10−4 = 1.1213325 x 10−3 = 1.12 x 10−3; Mr (ascorbic acid) = 176; Mass ascorbic acid in 250 cm3 = 10 x Mr x moles ascorbic acid = 10 x 176 x

1.1213325 x 10−3 (1) (= 1.97355); Percentage ascorbic acid in tablet 100 x mass ascorbic acid in 250 cm3 ÷ 2 = 100 x

10 x 176 x 1.1213325 x 10−3 ÷ 2 = 98.67726 = 98.7%; Percentage mass calculation 2

Amount thiosulphate = 0.0331 dm3 x 0.1 mol dm —3 = 0.00331 mol= amount of copper(II) ions in 25 cm3 portion;

∴ amount Cu = 10 x 0.00331= 0.0331 mol in total; ∴ mass Cu = 0.0331 mol x 63.5 g mol­1 = 2.102 g; ∴ % copper = (2.102 x 100)÷ 3.00= 70.1%; to 3 s.f. only;

Failure to fill up jet of burette some reagent used to fill the jet (which does not react with the iodine solution) and so the

titre is too high; and hence the percentage value would be too high;

Why using four tablets gives more accurate reading than using two Using larger mass reduces the percentage error / uncertainty (in weighing);

ALLOW: using four tablets gives a more representative sample; How to determine if a mixture is racemic or not

Use of (plane­)polarized light (mentioned somewhere); ALLOW Use a polarimeter;

Pure optical isomer / enantiomer) rotates the plane of (plane­) polarized light; ALLOW rotates plane­polarized light scores 2;

Questions on ester bonds and what happens to them in given scenario

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Peter Williams

(Ester group / vitamin C / it) is hydrolysed; IGNORE Just ‘breaks down’

Draw a polymer questions i) (draw extended polymer unit)

amide linkage; ALLOW CONH for amide linkage;

Completion of structure (brackets not required) with displayed or skeletal formula; ii) (type of polymerisation)

Condensation / addition­elimination (polymerisation); ii) (where this linkage naturally occurs)

Protein / proteins / polypeptide / polypeptides / peptide / peptides; ALLOW Enzyme / Enzymes;

Types of bonds between polymer molecules Hydrogen bonds; and (permanent) dipole(­permanent dipole) forces; and London/van der Waals’/dispersion forces;

Hydrogen bonding: (Between) the hydrogen atoms on the nitrogen atoms and .... (the lone pair of electrons

on) oxygen/ nitrogen atoms; Dipole­dipole (permanent) forces:

Permanent dipole­permanent dipole forces; Because the C=O / carbon­oxygen bond/the C­N bond is polar/a dipole; OR N and/or O are electronegative atoms;

London forces: Polymer has large number of/many electrons;

OR Explanation e.g temporary/induced/fluctuating dipoles;

Identify unknown IR+ NMR spectrum (scenario)

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Peter Williams

i) (Labelling Hydrogen environments)

Unique NH (at e); Unique CH2 (at c); CH (at d) and CH (at f) with different unique labels; 2CH2 (at a) and 2CH2 (at b) with different new labels;

ii) (IR spectrum consistent with structure) CO amide (stretching vibrations are in the region) 1700­1630 cm−1; N—H amide (stretching vibrations are in the region) 3500­3140 cm−1;

*note: focus on the main function groups with IR spectrum iiI) (cis not trans isomer)

In the trans isomer the (amine and acid chloride) groups are too far apart to react intramolecularly / to form M;

OR Because the groups are on opposite sides of the (plane of the) ring;

Why NMR of paracetamol contains 2 doublets and rest singlets The (ring) hydrogen atoms are on carbon atoms which have one / a hydrogen on an

adjacent carbon atom, so are doublets;

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Peter Williams

All the other hydrogen atoms have no adjacent hydrogen (bearing carbon) atoms, so are singlets;

Thiosulfate and Iodine (reaction) I2 + 2S2O32− → 2I− + S4O62−;

Steam distillation explanation + advantage over normal distillation steam is passed into the mixture;

OR water is added and mixture boiled or distilled or heated; and the 2­nitrophenol / product vapour distilled off with the water (and condensed);

Advantage: The 2­nitrophenol / product distils at a lower temperature / prevents decomposition;

Reflux apparatus (diagram) Pear shaped/round bottomed flask & heat source;

Allow vertical arrow with or without the word heat Allow water bath as a heat source

Liebig condenser, shown vertically; (Water) flow shown correctly into a jacket;

Ignore thermometers unless stoppered

Penalise (one for each): Stopper/sealed

Gaps between flask and condenser Condenser inner tube extends into liquid in flask

Reasons for heating under reflux Heat

Speed up reaction / to overcome the activation energy / provide energy to break bonds / because activation energy for the reaction is high;

Under reflux Prevent escape of reactants / products;

Or As they may be flammable / harmful / volatile;

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Peter Williams

Calculation of % yield Moles of 2­hydroxybenzoic acid = 9.4/138 (= 0.0681); So theoretical yield of aspirin = 0.0681 x 180 = 12.26 g; % yield = 100 x 7.77/12.26 = 63.4%;

Purifying a solid via recrystallization Dissolve/add to impure solid in min. volume / amount; of hot solvent / water; (Filter whilst hot) Allow to cool and filter off product / (re)crystallize and filter off product; Wash with cold / small amount of solvent / water (then dry);

Effect on yield/disadvantage of recrystallizing It reduces yield as some product remains in solution;

Catalysis process and how it speeds up a reaction Gases adsorb onto / bind to catalyst (surface);

Allow gases are absorbed onto surface; Then react and desorb / leave;

Reaction could be faster because: These processes lower the activation energy (by providing an alternative route so a

greater proportion of molecules react); Bonds in reactant(s) are weakened; Reactants may be positioned in more favourable orientations; Reactants can migrate towards each other on surface; Increases likelihood of molecules coming into contact / colliding; Adsorption onto surface means more reactant molecules in a given space;

How transition metals can act as catalysts Transition metals form various/variable oxidation states; They are able to donate and receive electrons/they are able to oxidize and reduce/they

are able to be oxidized and reduced /ions contain partially filled (sub­) shells of d electrons;

Disadvantage of using a hydrogen­oxygen fuel cell to power a motor Storage (problems);

OR hydrogen / oxygen / the gases have to be stored under pressure;

Explanation of breathalyser and how it shows % of ethanol present Early breathalysers: (the extent to which) dichromate turns green; Fuel cells: (more alcohol means larger)current / quantity of electricity; Infrared breathalysers: (more alcohol means greater) absorbance;

Why IR breathalysers do not detect ­OH group in breath Water (in the breath) also has an OH bond;

Why police use IR breathalysers together with fuel cell breathalysers Additional evidence is more reliable;

OR

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Peter Williams

Police often use IR as well as fuel cell breath test to provide sufficient evidence to prosecute (without need for blood test);

OR Fuel cell breathalysers are portable and determine whether or not to check with IR at the

police station; Dot & cross arrangement of electrons in CN­

Accept dots, crosses, mixture of both; Triple bond; Non­bonding electrons;

IGNORE presence/absence of negative charge But if positive charge max 1 Cyanide acting as a ligand

The non­bonding / lone pair of electrons on the carbon; ALLOW non­bonding/lone pair of electrons on the nitrogen;

Forms a dative covalent/coordinate bond (to central metal ion); Features of a base (amine from text)

Lone pair (of electrons on the nitrogen atom); ALLOW non­bonded pair (of electrons on the nitrogen atom);

Why minimum amount of solvent is used (Otherwise) too much (product) remains in solution;

OR If excess (solvent) is used, crystals might not form;

ALLOW:To avoid losing (too much) product (in the filtrate when crystallization occurs) /

‘to maximize the yield’/ ‘will crystallize better from a concentrated solution’/ ‘will recrystallize (better) when cold’;

Removing insoluble impurities By hot filtration / During the first filtration / During the second step in the process;

Removing soluble impurities By remaining in solution / Left in filtrate / Removed when washed (with cold solvent);

Testing for purity after recrystallization Measure the melting temperature / melting point; compare with data / known value (from a data book / literature / Internet /data base)

(BOTH points needed for the mark); Lattice melting point values

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Peter Williams

TiO2 ‘Structure’ mark Giant (structure);

OR Lattice (structure);

TiO2 ‘Bonding’ mark Strong (electrostatic) attraction between ions;

ALLOW:Strong ionic bonds / ionic bonds require a lot of energy to break; TiCl4 ‘Structure’ mark

(Simple) molecules / (small) molecules /molecular; TiCl4 ‘Bonding’ mark

Weak London / dispersion / van der Waals’ forces (between molecules) / London /dispersion / van der Waals’ forces (between molecules) require little energy to break;

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